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The answer is in terms of k. I tried finding a pattern between the consecutive terms of the series but could find none. Also I feel like there might be a systematic way for solving this that I do not know. My question is how would you solve this problem?

Thank you!

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I presume that the Wikepedia site Ross Milikan links to says this: If we try a solution of the form $a_k= a^k$, for some constant k, then $a_{k-1}= a^{k-1}= \frac{a^k}{a}$ and $a_{k-2}= a^{k-2}= \frac{a^k}{a^2}$ so that $a_k- 4a_{k-1}+ 3a_{k-2}= a^k- \frac{4a^k}{a}+ \frac{4a^k}{k^2}= 0$. Dividing by $a^k$ (which, for a non-zero, is never 0) we have $1- \frac{4}{a}+ \frac{3}{a^2}= 0$. Multiplying by $a^2$ this becomes $a^2- 4a+ 3= (a- 3)(a- 1)= 0$ which has roots for a= 3 and a= 1. So both $a_k= 3^k$ and $a_k= 1^k= 1$ are solutions. And it is then easy so show that $a_k= C_13^k+ C_2$, where $C_1$ and $C_2$ can be any constants, is the "general" solution.

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  • $\begingroup$ I believe you mean "for some constant $a$". $\endgroup$ – YoTengoUnLCD Jul 15 '16 at 21:40
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Let $A(x)=\sum\limits_{k=1}^{\infty}a_kx^{k}$ be the generating function of the sequense $\{a_k\}$.

Now, by summing up from $k=3$ to $n$ the recursive relation we have that

$$ \sum\limits_{k=3}^{n}a_kx^k-4\sum\limits_{k=3}^{n}a_{k-1}x^k+3\sum\limits_{k=3}^{n}a_{k-2}x^k=0\Longrightarrow \sum\limits_{k=3}^{n}a_kx^k-4x\sum\limits_{k=2}^{n-1}a_{k}x^k+3x^2\sum\limits_{k=1}^{n-2}a_{k}=0 $$

Let $n$ go to $\infty$

$$ \sum\limits_{k=3}^{\infty}a_kx^k-4x\sum\limits_{k=2}^{\infty}a_{k}x^k+3x^2\sum\limits_{k=1}^{\infty}a_{k}=0\Longrightarrow [A(x)-a_2x^2-a_1x]-4x[A(x)-a_1x]+3x^2A(x)=0\Longrightarrow [A(x)-2x^2-x]-4x[A(x)-x]+3x^2A(x)=0\Longrightarrow (3x^2-4x+1)A(x)=2x^2+2x\Longrightarrow A(x)=\cfrac{2x^2+2x}{3x^2-4x+1} $$

and so by partially decomposition the fraction $\cfrac{2x^2+2x}{3x^2-4x+1}$ we have that

$$A(x)=\cfrac{1}{1-3x}-\cfrac{1}{1-x}$$

The generating functions of $\cfrac{1}{1-3x}$ and $\cfrac{1}{1-x}$ is $\sum\limits_{k=1}^{\infty}3^kx^k$ and $\sum\limits_{k=1}^{\infty}x^k$

Hence that,

$$A(x)=\sum\limits_{k=1}^{\infty}3^kx^k + \sum\limits_{k=1}^{\infty}x^k=\sum\limits_{k=1}^{\infty}(3^k+1)x^k$$

So, $ \ \ \ a_k=3^k+1$, $\ \ k\in\Bbb N$

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